如何根据表扫描每个单词值，然后对其进行计算并从中制作VSM(向量空间模型)

Question

说我有一个表，其中包含另一个表中每个单词的概率。该表有2个类；实际和 non_actual 。我将命名 master_table

 actual = [0.5;0.4;0.6;0.75;0.23;0.96;0.532]; %sum of the probabilities is 1.     
actual + non_actual = 1
non_actual = [0.5;0.6:0.4;0.25;0.77;0.04;0.468];
words = {'finn';'jake';'iceking';'marceline';'shelby';'bmo';'naptr'};
master_table = table(actual,non_actual,...
'RowNames',words)

然后我有一个包含句子的表。我将命名为T2

sentence = {'finn marceline naptr';'jake finn simon marceline haha';'jake finn finn jake iceking';'bmo shelby shelby finn naptr';'naptr naptr jake finn bmo shelby'}
T2 = table('RowNames',sentence)

如何制作这样的方法（不属于" simon"，" haha"的单词不属于主人，因此它不会影响确定类的概率的计算）：

                                    actual %determines the value based on probabilities from each words%        non_actual               class
finn marceline naptr                0.5 * 0.75 * 0.532                                                         0.5 * 0.25 * 0.468        compares the value from each class. if actual > non_actual then the class should be "actual"
jake finn simon marceline haha      0.4 * 0.5 * 1 * 0.25 * 1                                                   0.6 * 0.5 * 1 * 0.75 * 1
jake finn finn jake iceking
bmo shelby shelby finn naptr
naptr naptr jake finn bmo shelby

以及如何从上面的问题中制作VSM（向量空间模型）：

                                                                        WORDS                                   
                                    | bmo | finn | jake | iceking | haha | marceline | naptr | shelby | simon |     %words sorted alphabetically      
finn marceline naptr                   0     1       0        0       0        1         1       0       0      
jake finn simon marceline haha         0     1       1        0       1        1         0       0       1
jake finn finn jake iceking            0     2       2        1       0        0         0       0       0
bmo shelby shelby finn naptr           1     1       0        0       0        0         1       1       0      
naptr naptr jake finn bmo shelby       1     1       1        0       0        0         1       1       0       

Answer 1

作为一个快速解决方案，以下代码应解决您的问题：

% Split the sentence into single strings
s = strsplit(sentence{1});
% loop over all single strings
for i=1:length(s)
    % search for each string pattern in the words-cell
    c = strfind(words,s{i});
    % get a logical vector for getting the index of the found pattern in
    % the words-cell
    ix=cellfun('isempty', c);
    ind = find(ix == 0);
    if actual(ind) > non_actual(ind)
        % do something with actual...
    end;
end;

您应该阅读代码中使用的每个功能的帮助章节：strsplit strfind cellfun，以获取有关其工作方式的更多信息。

Answer 2

这也有点循环，但我像性能一样跌倒不是问题。我将首先创建一个较大的表，然后更改循环中的值：

T2 = table(ones(height(T2),1),ones(height(T2),1),repmat({''},height(T2),1),'RowNames',sentence,'VariableNames',{'actual' 'non_actual' 'outcome'});
for i=1:height(T2)
    % split the row name
    A=strsplit([T2.Properties.RowNames{i,:}]);
    actual=1; %which is neutral for multiplication
    non_actual=1; 
    for j=1:length(A)
       actual = actual *  master_table{A(j),1};
       non_actual = non_actual *  master_table{A(j),2};
    end
    %if you need those
    T2.actual(i)=actual;
    T2.non_actual(i)=non_actual;
    if actual > non_actual
        T2.outcome(i)={'actual'};
    else
        T2.outcome(i)={'non_actual'};
    end;
end;